AmirG - Challanges completed succesfully

[amirg33]

No description provided.

[bripollc]

Hi Amir,

Congratulations!!!!!!!!! I have not much to tell you because you have made a really good second lab:)

✨Challenge 1: everything correct, it's perfect!

✨Challenge 2: idem, good job.
The only thing:

1. Using the pop method, remove the first element of set1.

set1_list = list(set1)
set1_list.pop(0)
set1_list_pop = set(set1_list) # And we convert it back to set
print(set1_list_pop)

The pop method removes an element randomly, so we could convert it to list to pop the first element.

✨Challenge 3: nothing to say, it's perfect!

keep going!!!!! 🚀

[amirg33]

Perfect! 😊 Thank you for your feedback! I have a question though. In the Challange 3 we had to sort the list. I used the .sort () method, but my first approach was to convert the list into a set and back into a list in order to sort it. But it did not work. I tried to solve the problema yesterday with Fer but we weren’t able to understand why the code does not work. Thank you in advanced. Attached my idea: keys = list(word_freq) set(list(keys)) word_freq2 = {} for i in keys: word_freq2[i] = word_freq.get(i) print(word_freq2) word_freq == word_freq2 Best Regards Amir De: ***@***.***> Enviado: miércoles, 11 de octubre de 2023 21:03 Para: ***@***.***> CC: Amir ***@***.***>; ***@***.***> Asunto: Re: [Ironhack-data-bcn-oct-2023/lab-dict-sets-tuple] AmirG - Challanges completed succesfully (PR #13) Hi Amir, Congratulations!!!!!!!!! I have not much to tell you because you have made a really good second lab:) ✨Challenge 1: everything correct, it's perfect! ✨Challenge 2: idem, good job. The only thing: 1. Using the pop method, remove the first element of set1. set1_list = list(set1) set1_list.pop(0) set1_list_pop = set(set1_list) # And we convert it back to set print(set1_list_pop) The pop method removes an element randomly, so we could convert it to list to pop the first element. ✨Challenge 3: nothing to say, it's perfect! keep going!!!!! 🚀 — Reply to this email directly, view it on GitHub<#13 (comment)>, or unsubscribe<https://github.com/notifications/unsubscribe-auth/BCXWFKQ6APAEQFYSNEWQCXDX63UQ5ANCNFSM6AAAAAA52WS3II>. You are receiving this because you authored the thread.Message ID: ***@***.***>